Question

如图,在△ABC中,∠BAC=90°,AB=AC=6,D为BC的中点

Answer 1

∠EDF=90°因为∠BAC=90°,AB=AC,D是BC的中点,AE=BF,连接AD,可证明△DAE≌△DBF,则有DE=DF,再用角与角之间的关系求得∠EDF是直角,即可判断△DEF为等腰直角三角形．连接AD,∵Rt△ABC中,∠BAC=90°,AB=AC,∴∠B=∠C=45°．∵AB=AC,DB=CD,∴∠DAE=∠BAD=45°．∴∠BAD=∠B=45°．∴AD=BD,∠ADB=90°．∵AE=BF,∠DAE=∠B=45°,AD=BD,∴△DAE≌△DBF（SAS）．∴DE=DF,∠ADE=∠BDF．∵∠BDF+∠ADF=∠ADB=90°,∴∠ADE+∠ADF=90°．